Problem: Solve
\[\arccos 2x - \arccos x = \frac{\pi}{3}.\]Enter all the solutions, separated by commas.
From the given equation,
\[\arccos 2x = \arccos x + \frac{\pi}{3}.\]Then
\[\cos (\arccos 2x) = \cos \left( \arccos x + \frac{\pi}{3} \right).\]Hence, from the angle addition formula,
\begin{align*}
2x &= \cos (\arccos x) \cos \frac{\pi}{3} - \sin (\arccos x) \sin \frac{\pi}{3} \\
&= \frac{x}{2} - \frac{\sqrt{3}}{2} \sqrt{1 - x^2},
\end{align*}so
\[-3x = \sqrt{3} \cdot \sqrt{1 - x^2}.\]Squaring both sides, we get $9x^2 = 3 - 3x^2.$  Then $12x^2 = 3,$ so $x^2 = \frac{1}{4},$ and $x = \pm \frac{1}{2}.$  Checking, we find only $x = \boxed{-\frac{1}{2}}$ works.